If X is a non-negative random variable for which E(X) exists,show that for every t>0.
$\displaystyle P[X\geq t] \leq\frac {E(X)}{t}$
$\displaystyle \frac{E(X)}{t}=\int_0^{\infty} \frac{x}{t}p(x)\; dx = \int_0^t\frac{x}{t}p(x)\; dx + \int_t^{\infty}\frac{x}{t}p(x)\; dx$
.........$\displaystyle \ge \int_t^{\infty}\frac{x}{t}p(x)\; dx$
and as $\displaystyle \frac{x}{t}\ge 1$ for $\displaystyle x\ge t$ hence:
$\displaystyle \frac{E(X)}{t}\ge \int_t^{\infty}\frac{x}{t}p(x)\; dx\ge \int_t^{\infty}p(x)\; dx$
CB
And you didn't notice that under Mr Fantasy's link...
http://people.csail.mit.edu/ronitt/COURSE/S07/lec25.pdf
under your last post...
http://www.mathhelpforum.com/math-he...nequality.html
it's Theorem ONE, on page ONE.